Problem: Let $a,$ $b,$ $c,$ $d$ be positive integers such that
\[\begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 18 & 12 \\ -20 & -13 \end{pmatrix}.\]Find the smallest possible value of $a + b + c + d.$
Answer: Performing the multiplication on both sides, we obtain
\[\begin{pmatrix} 3a & 3b \\ 2c & 2d \end{pmatrix} = \begin{pmatrix} 18a - 20b & 12a - 13b \\ 18c - 20d & 12c - 13d \end{pmatrix}.\]Hence, $3a = 18a - 20b,$ $12a - 13b = 3b,$ $18c - 20d = 2c,$ and $12c - 13d = 2d.$  Then $15a = 20b,$ $12a = 16b,$ $16c = 20d,$ and $12c = 15d.$  These reduce to $3a = 4b$ and $4c = 5d.$  The smallest positive integer solutions are $a = 4,$ $b = 3,$ $c = 5,$ and $d = 4,$ so the smallest possible value of $a + b + c + d$ is $4 + 3 + 5 + 4 = \boxed{16}.$